Solve 10m^2+35m+18m-63=0

$$10 m ^ { 2 } + 35 m + 18 m - 63 = 0$$

Solve for m

$$m = -\frac{63}{10} = -6\frac{3}{10} = -6.3$$

$$m=1$$

Steps Using Factoring By Grouping

Combine 35m and 18m to get 53m.

$$10m^{2}+53m-63=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 10m^{2}+am+bm-63. To find a and b, set up a system to be solved.

$$a+b=53 ab=10\left(-63\right)=-630$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -630.

$$-1,630 -2,315 -3,210 -5,126 -6,105 -7,90 -9,70 -10,63 -14,45 -15,42 -18,35 -21,30$$

Calculate the sum for each pair.

$$-1+630=629 -2+315=313 -3+210=207 -5+126=121 -6+105=99 -7+90=83 -9+70=61 -10+63=53 -14+45=31 -15+42=27 -18+35=17 -21+30=9$$

The solution is the pair that gives sum 53.

$$a=-10 b=63$$

Rewrite 10m^{2}+53m-63 as \left(10m^{2}-10m\right)+\left(63m-63\right).

$$\left(10m^{2}-10m\right)+\left(63m-63\right)$$

Factor out 10m in the first and 63 in the second group.

$$10m\left(m-1\right)+63\left(m-1\right)$$

Factor out common term m-1 by using distributive property.

$$\left(m-1\right)\left(10m+63\right)$$

To find equation solutions, solve m-1=0 and 10m+63=0.

$$m=1 m=-\frac{63}{10}$$

Steps Using the Quadratic Formula

Combine 35m and 18m to get 53m.

$$10m^{2}+53m-63=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, 53 for b, and -63 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-53±\sqrt{53^{2}-4\times 10\left(-63\right)}}{2\times 10}$$

Square 53.

$$m=\frac{-53±\sqrt{2809-4\times 10\left(-63\right)}}{2\times 10}$$

Multiply -4 times 10.

$$m=\frac{-53±\sqrt{2809-40\left(-63\right)}}{2\times 10}$$

Multiply -40 times -63.

$$m=\frac{-53±\sqrt{2809+2520}}{2\times 10}$$

Add 2809 to 2520.

$$m=\frac{-53±\sqrt{5329}}{2\times 10}$$

Take the square root of 5329.

$$m=\frac{-53±73}{2\times 10}$$

Multiply 2 times 10.

$$m=\frac{-53±73}{20}$$

Now solve the equation m=\frac{-53±73}{20} when ± is plus. Add -53 to 73.

$$m=\frac{20}{20}$$

Divide 20 by 20.

$$m=1$$

Now solve the equation m=\frac{-53±73}{20} when ± is minus. Subtract 73 from -53.

$$m=-\frac{126}{20}$$

Reduce the fraction \frac{-126}{20} to lowest terms by extracting and canceling out 2.

$$m=-\frac{63}{10}$$

The equation is now solved.

$$m=1 m=-\frac{63}{10}$$

Steps for Completing the Square

Combine 35m and 18m to get 53m.

$$10m^{2}+53m-63=0$$

Add 63 to both sides. Anything plus zero gives itself.

$$10m^{2}+53m=63$$

Divide both sides by 10.

$$\frac{10m^{2}+53m}{10}=\frac{63}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$m^{2}+\frac{53}{10}m=\frac{63}{10}$$

Divide \frac{53}{10}, the coefficient of the x term, by 2 to get \frac{53}{20}. Then add the square of \frac{53}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m^{2}+\frac{53}{10}m+\left(\frac{53}{20}\right)^{2}=\frac{63}{10}+\left(\frac{53}{20}\right)^{2}$$

Square \frac{53}{20} by squaring both the numerator and the denominator of the fraction.

$$m^{2}+\frac{53}{10}m+\frac{2809}{400}=\frac{63}{10}+\frac{2809}{400}$$

Add \frac{63}{10} to \frac{2809}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$m^{2}+\frac{53}{10}m+\frac{2809}{400}=\frac{5329}{400}$$

Factor m^{2}+\frac{53}{10}m+\frac{2809}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(m+\frac{53}{20}\right)^{2}=\frac{5329}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m+\frac{53}{20}\right)^{2}}=\sqrt{\frac{5329}{400}}$$

Simplify.

$$m+\frac{53}{20}=\frac{73}{20} m+\frac{53}{20}=-\frac{73}{20}$$

Subtract \frac{53}{20} from both sides of the equation.

$$m=1 m=-\frac{63}{10}$$